If the line `y=mx`, meets the lines `x+2y=1` and `2x-y+3=0` at one point only then `m=?`

Options

A. `1`
B. `-1`
C. `-2`
D. None of these

Answer

To solve the problem, we need to find the value of \( m \) such that the line \( y = mx \) intersects the lines \( x + 2y = 1 \) and \( 2x - y + 3 = 0 \) at only one point. This means that the line \( y = mx \) must pass through the intersection point of the two given lines.

### Step 1: Find the intersection point of the lines \( x + 2y = 1 \) and \( 2x - y + 3 = 0 \).

We will solve these two equations simultaneously.

1. **Equation 1:** \( x + 2y = 1 \)
2. **Equation 2:** \( 2x - y + 3 = 0 \) or \( 2x - y = -3 \)

We can express \( y \) from Equation 1:
\[ 2y = 1 - x \implies y = \frac{1 - x}{2} \]

Now substitute \( y \) into Equation 2:
\[ 2x - \frac{1 - x}{2} = -3 \]

### Step 2: Clear the fraction by multiplying the entire equation by 2:
\[ 4x - (1 - x) = -6 \]
\[ 4x - 1 + x = -6 \]
\[ 5x - 1 = -6 \]
\[ 5x = -5 \implies x = -1 \]

### Step 3: Substitute \( x = -1 \) back into Equation 1 to find \( y \):
\[ -1 + 2y = 1 \implies 2y = 2 \implies y = 1 \]

Thus, the intersection point \( P \) of the two lines is:
\[ P(-1, 1) \]

### Step 4: Since the line \( y = mx \) must pass through the point \( P(-1, 1) \), we substitute this point into the equation:
\[ y = mx \implies 1 = m(-1) \]
\[ 1 = -m \implies m = -1 \]

### Conclusion:
The value of \( m \) is:
\[ \boxed{-1} \]